Calculating Charge on Parallel Plates

AI Thread Summary
An electron accelerates between charged parallel plates, starting with a velocity of 1.63 x 10^4 m/s and reaching 4.15 x 10^4 m/s over a distance of 2.10 cm. The discussion focuses on calculating the charge on the plates, with the user initially struggling with the equations and concepts involved. Key equations include E = (4πkQ)/A and Q = CV, where voltage (V) must be determined from the change in kinetic energy. After several calculations and corrections, the user successfully finds the charge on each plate to be approximately 1.27 x 10^-13 C, confirming the correct answer after extensive troubleshooting. The conversation highlights the importance of understanding energy conservation in electric fields and the relationships between charge, capacitance, and voltage.
Etopn23
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Homework Statement


An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63 x 10^4 m/s to the right. Its speed on reaching the other plate, 2.10 cm away is 4.15x10^4 m/s

I got parts a and b - part C is what I'm having problems with.

C) If the plates are square with an edge length of 25.4 cm, determine the charge on each.


Homework Equations


E=(4πkQ)/A


The Attempt at a Solution



I substituted for all the unknowns except for E and Q - but that leaves me with two unknowns; I'm completely flabbergasted with what to do next.
 
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Can you post what parts a and b said, and what your answers were?
 
A) What type of charge is on each plate?
Left Plate Negative, Right Plate positive

B) What is the direction of the electric field between the plates?
From right to left
 
For a capacitor Q = CV.

Try to calculate C and V... then you can get Q.
 
learningphysics said:
For a capacitor Q = CV.

Try to calculate C and V... then you can get Q.
Thanks a lot - i'll go do that. I didn't have that equation unfortunately xD.
 
Ok I plugged in

C=((8.85x10^-12)(9x10^9)(.064516))/(2.10x10^-2)=.2447

Then Q=(.2447)(1.63X10^4)= 3988.61

Q=(.2447)(4.15x10^4)= 10155

I plugged each Q into E=(4pikQ)/A and I got two answers that were a bit off =\ No idea what I did wrong.

I know the final answer is 1.13 x 10^-13 but I'm not getting anywhere near that.
 
Etopn23 said:
Ok I plugged in

C=((8.85x10^-12)(9x10^9)(.064516))/(2.10x10^-2)=.2447

Then Q=(.2447)(1.63X10^4)= 3988.61

Q=(.2447)(4.15x10^4)= 10155

I plugged each Q into E=(4pikQ)/A and I got two answers that were a bit off =\ No idea what I did wrong.

I know the final answer is 1.13 x 10^-13 but I'm not getting anywhere near that.

The V in Q = CV is voltage not speed. You need to find the voltage between the two plates of the capacitor... use energy to find voltage...
 
learningphysics said:
The V in Q = CV is voltage not speed. You need to find the voltage between the two plates of the capacitor... use energy to find voltage...

Aaah sorry; just assumed what the variables were since I hadn't encountered this equation before - let me fix this now xD.
 
Ok so V=Ue/Qe

Ue= qeEd but E is unknown =\.

I'm reading up on this equation (surprisingly in my book it isn't in the chapter that the problem was assigned from - it's in the next chapter) so I haven't had exposure to it.

Am I doing something wrong? =\
 
  • #10
Etopn23 said:
Ok so V=Ue/Qe

Ue= qeEd but E is unknown =\.

But you know that V = Ed...

Ue = qe*V
 
  • #11
Unless I'm missing something (which I problem am due to my own ignorance ;_;) I still have multiple unknowns.

UGH, it's irritating having an assignment which seems easy until you get to the last problem.

Any other clues you can prompt me to? =\
 
  • #12
Etopn23 said:
Unless I'm missing something (which I problem am due to my own ignorance ;_;) I still have multiple unknowns.

UGH, it's irritating having an assignment which seems easy until you get to the last problem.

Any other clues you can prompt me to? =\

What is the change in kinetic energy of the electron? electric potential energy is being converted to kinetic energy...
 
  • #13
learningphysics said:
What is the change in kinetic energy of the electron? electric potential energy is being converted to kinetic energy...

If I calculated it correctly I got 1.147x10^-26 - plugged that in for delta Ue; plugged Ue into the V equation - plugged V into the Q equation - which I plugged in for E and I got the wrong answer =s.
 
  • #14
Etopn23 said:
If I calculated it correctly I got 1.147x10^-26 - plugged that in for delta Ue; plugged Ue into the V equation - plugged V into the Q equation - which I plugged in for E and I got the wrong answer =s.

That's not what I'm getting for change in kinetic energy... can you show your calculations?
 
  • #15
learningphysics said:
That's not what I'm getting for change in kinetic energy... can you show your calculations?

KEo= (1/2)(9.10938188 × 10-31)(1.63x10^4)
KEf= (1/2)(9.10938188 × 10-31)(4.15x10^4)

KEf-KEo=Delta KE which is where I got 1.148E-26

I assumed the change in kinetic energy would be equal to the change in potential energy. So I took the magnitude of it and plugged in all the way through - lead me to a V of 7.17334E-8
which then lead me to a Q of 1.7E-8 which I proceeded to plug into the initial equation; I got a number that I knew was off immediately so I didn't bother writing it down =\.
 
  • #16
Etopn23 said:
KEo= (1/2)(9.10938188 × 10-31)(1.63x10^4)
KEf= (1/2)(9.10938188 × 10-31)(4.15x10^4)

KEf-KEo=Delta KE which is where I got 1.148E-26

I assumed the change in kinetic energy would be equal to the change in potential energy. So I took the magnitude of it and plugged in all the way through - lead me to a V of 7.17334E-8
which then lead me to a Q of 1.7E-8 which I proceeded to plug into the initial equation; I got a number that I knew was off immediately so I didn't bother writing it down =\.

you didn't square the velocties... (1/2) mv^2...
 
  • #17
learningphysics said:
you didn't square the velocties... (1/2) mv^2...
Ouch >.<. K - correcting my mistake.
 
  • #18
Etopn23 said:
Ok I plugged in

C=((8.85x10^-12)(9x10^9)(.064516))/(2.10x10^-2)=.2447

also here... Capacitance = epsilon*A/d

I think the k is a mistake.
 
  • #19
Ok corrected; that got my Ue to 6.55452E18 - which when plugged into my V got me 4.09658E37. That lead to Q being 1.00243E37 and E being some huge number =\.
 
  • #20
learningphysics said:
also here... Capacitance = epsilon*A/d

I think the k is a mistake.

aaah ok i'll check on that - thanks; I knew something had to be wrong
 
  • #21
Ended up changing my C to 1.29E-9 =s still got a massive number for E when it was all said and done.
 
  • #22
Etopn23 said:
Ended up changing my C to 1.29E-9 =s still got a massive number for E when it was all said and done.

Do you need to calculate E? Does the question ask for it?

I'm getting U = 6.627E-22
 
  • #23
learningphysics said:
Do you need to calculate E? Does the question ask for it?

I'm getting U = 6.627E-22
Omg I am an IDIOT I used K instead of M for no apparent reason when calculating it unconsciously... Sorry let me rework this =\
 
  • #24
I got 6.6339E-22 for U, .004146 for E and 5.36811E-12.

I know it's asking for the charge on each plate - but I have no idea how to find that since I've never had a problem like this. I was planning on finding E then plugging it into E=(Kq1)/(r1)^2 and use the midpoint between the two plates as r. =s Iunno if that's the correct thing to do though.
 
  • #25
Etopn23 said:
I got 6.6339E-22 for U, .004146 for E and 5.36811E-12.

I know it's asking for the charge on each plate - but I have no idea how to find that since I've never had a problem like this. I was planning on finding E then plugging it into E=(Kq1)/(r1)^2 and use the midpoint between the two plates as r. =s Iunno if that's the correct thing to do though.

You got U... So calculate voltage using U = qe*V... Then using Q = CV... calculate Q on the capacitor plate... what did you get for C?
 
  • #26
Etopn23 said:
KEo= (1/2)(9.10938188 × 10-31)(1.63x10^4)
KEf= (1/2)(9.10938188 × 10-31)(4.15x10^4)

KEf-KEo=Delta KE which is where I got 1.148E-26

I assumed the change in kinetic energy would be equal to the change in potential energy. So I took the magnitude of it and plugged in all the way through - lead me to a V of 7.17334E-8
which then lead me to a Q of 1.7E-8 which I proceeded to plug into the initial equation; I got a number that I knew was off immediately so I didn't bother writing it down =\.

I just did this and got 1.15x10^-13 - I know the correct answer is 1.13x10^-13 O_O

*edit* Meant to quote what you just Quoted, I have no idea why it quoted this.
 
Last edited:
  • #27
learningphysics said:
You got U... So calculate voltage using U = qe*V... Then using Q = CV... calculate Q on the capacitor plate... what did you get for C?
1.29471E-9
 
  • #28
Etopn23 said:
1.29471E-9

I'm not getting that. Did you get V from U = qe*V ?
 
  • #29
learningphysics said:
I'm not getting that. Did you get V from U = qe*V ?

Yeah - 6.6339E-33/1.60x10^-19= .004146

Then I plugged .004146 into Q=VC; (.004146 )(1.29471E-9) and I got 5.36811E-12 for Q
 
  • #30
Etopn23 said:
Yeah - 6.6339E-33/1.60x10^-19= .004146

Then I plugged .004146 into Q=VC; (.004146 )(1.29471E-9) and I got 5.36811E-12 for Q

The 1.29E-9 is wrong for capacitance...
 
  • #31
learningphysics said:
The 1.29E-9 is wrong for capacitance...
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.
 
  • #32
Etopn23 said:
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.

:smile: You actually meant 1.127E-13 right?
 
  • #33
Etopn23 said:
AAAH. Fixed that and plugged in again.
YES I got 1.27E-13 for the magnitude of the plates. WHICH IS THE CORRECT ANSWER :D. OMG. It only took 4 hours xD.

I just wanted to thank you for being tenacious enough to stick with the help to the end. I'm so slow with new concepts ><. Thanks alot, I really appreciate it.

no prob. :smile: everything takes time, experience, practice etc... just keep doing the problems, and it will become easier...
 
  • #34
learningphysics said:
:smile: You actually meant 1.127E-13 right?
Yeah - Typo sorry xD
 
  • #35
learningphysics said:
no prob. :smile: everything takes time, experience, practice etc... just keep doing the problems, and it will become easier...
Oh I will O_O. Having this many problems on something is unacceptable =\.
 

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